 |
|
| Browse | Ask | Answer | Search | Join/Login |
|
|
||||
|
I'll help you get started on this. For them to sit alternately the first one in line must be a boy. There are 7 children and 4 are boys, so the probability that a boy is first is 4/7. The second person in line must be a girl, and if a boy was first then there are 3 girls out of 6 children remaining, so the probability of the 2nd child being a girl is 3/6. Hence the probability that the first child is a boy and the second child is a girl is 4/7 x 3/6. Now continue with this line of thinking to fill the remaining 5 slots. Post back your answer and we'll check it for you.
|
||||
| Question Tools | Search this Question |
Add your answer here.
Check out some similar questions!
Probablity are .22, 36, 28, 14 that an investor wl be able to earn profit for rs 25000,15000,5000,or loss rs 5000 wt is expacted probablity
If the probablity of a team winning any games is 2 out of 3, What is the probablity that the team would win exactly four games in a five game series
Here is another question I find difficult to comprehend for me... If Pr(A)=0.7, Pr (A Intersection B)=0.32 and the Pr(A` Intersection B`)=0.10, then the Pr (B)= I know the formula Pr(A Union B)=Pr(A)+Pr(B)-Pr(A Intersection B) but don't know what the dashes mean and how to work out Pr(B)
A card is selected at random from a standard deck of cards.Determine each probablity. P(5), P(red 10) , P(odd Number) , P(queen of hearts or jack of diamonds). :confused: :confused: :confused:
View more questions Search
|
